2026?27 Syllabus Focus
The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.
- Rates of change and increasing or decreasing functions.
- Maxima and minima using the first- and second-derivative tests.
- Simple problems that apply these ideas, including real-life situations.
Last 24-Hour Checklist
Rate of Change
- Basic: $\frac{dy}{dx}$ = rate of change of $y$ w.r.t. $x$
- Chain Rule: $\frac{dy}{dx} = \frac{dy}{dt} · \frac{dt}{dx}$
- Volume/Area formulas for sphere, cone, cylinder
- Marginal cost = $\frac{dC}{dx}$, Marginal revenue = $\frac{dR}{dx}$
Increasing/Decreasing
- Increasing: $f'(x) > 0$
- Decreasing: $f'(x) < 0$
- Find critical points: $f'(x) = 0$
- Test intervals on number line
Maxima/Minima Tests
- 1st Derivative Test: Sign change of $f'(x)$
- 2nd Derivative Test: $f'(c) = 0$ and check $f''(c)$
- $f''(c) < 0$ ? Local maxima
- $f''(c) > 0$ ? Local minima
Absolute Max/Min
- Find all critical points in $[a,b]$
- Evaluate $f$ at critical points
- Evaluate $f$ at end points $a$ and $b$
- Compare values to find max/min
Rate of Change (The Speedometer)
Think of a speedometer. It shows your speed at that exact moment.
$\frac{dy}{dx}$ is simply the instantaneous speed of $y$ changing with $x$.
Answer: $dA/dr = 2π r$. At $r=5$, Rate = $10π$. with respect to $x$.
If $y = f(x)$, then $\frac{dy}{dx}\bigg|_{x=x_0}$ represents the instantaneous rate of change of $y$ at $x = x_0$.
Chain Rule for Related Rates
Important Formulas (Must Remember!)
Circle
- Area: $A = π r^2$
- Circumference: $C = 2π r$
- $\frac{dA}{dr} = 2π r$
Sphere
- Volume: $V = \frac{4}{3}π r^3$
- Surface Area: $S = 4π r^2$
- $\frac{dV}{dr} = 4π r^2$
Cube
- Volume: $V = x^3$
- Surface Area: $S = 6x^2$
- $\frac{dV}{dx} = 3x^2$
Cone
- Volume: $V = \frac{1}{3}π r^2 h$
- Curved Surface: $S = π r l$
- $l = \sqrt{r^2 + h^2}$
Common Applications
- Marginal Cost: $MC = \frac{dC}{dx}$ (rate of change of total cost)
- Marginal Revenue: $MR = \frac{dR}{dx}$ (rate of change of total revenue)
- Marginal Profit: $MP = MR - MC$
Increasing & Decreasing Functions
- $f$ is increasing on $I$ if $x_1 < x_2$ in $I \Rightarrow f(x_1) \leq f(x_2)$
- $f$ is decreasing on $I$ if $x_1 < x_2$ in $I \Rightarrow f(x_1) \geq f(x_2)$
- $f$ is strictly increasing if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$
- $f$ is strictly decreasing if $x_1 < x_2 \Rightarrow f(x_1)> f(x_2)$
First Derivative Test for Monotonicity
- $f$ is increasing on $[a,b]$ if $f'(x) \geq 0$ for all $x \in (a,b)$
- $f$ is decreasing on $[a,b]$ if $f'(x) \leq 0$ for all $x \in (a,b)$
- $f$ is constant on $[a,b]$ if $f'(x) = 0$ for all $x \in (a,b)$
Strictly Increasing: $f'(x) > 0$
Tangents make acute angles ($θ < 90^°$)
Strictly Decreasing: $f'(x) < 0$
Tangents make obtuse angles ($θ > 90^°$)
Step-by-Step Working Rule
Step 1: Find $f'(x)$
Step 2: Solve $f'(x) = 0$ to find critical points
Step 3: Plot critical points on number line to divide into intervals
Step 4: Check sign of $f'(x)$ in each interval (pick test points)
Step 5: If $f'(x) > 0 \implies$ increasing; if $f'(x) < 0 \implies$ decreasing
Example: If $f'(x) = (x-1)(x+2)(x-3)$, check signs at $x = -3, 0, 2, 4$:
- At $x = -3$: $(-)(-)(-) = - $ ? Decreasing
- At $x = 0$: $(-)(+)(-) = +$ ? Increasing
- At $x = 2$: $(+)(+)(-) = -$ ? Decreasing
- At $x = 4$: $(+)(+)(+) = +$ ? Increasing
Maxima and Minima
Key Definitions
Local Minima: $c$ is a point of local minima if there exists $h > 0$ such that: $$f(c) \leq f(x) for all x \in (c-h, c+h)$$
Critical Point: A point $c$ where either $f'(c) = 0$ or $f$ is not differentiable.
First Derivative Test
| Sign of $f'(x)$ as $x$ increases through $c$ | Nature at $c$ |
|---|---|
| Changes from Positive (+) to Negative (-) | Local Maxima (Graph goes up, then down) |
| Changes from Negative (-) to Positive (+) | Local Minima (Graph goes down, then up) |
| Does not change sign | Point of Inflection |
Second Derivative Test
- If $f''(c) < 0$ $\implies$ $c$ is point of local maxima.
- If $f''(c) > 0$ $\implies$ $c$ is point of local minima.
- If $f''(c) = 0$ $\implies$ Test fails. Use the First Derivative Test.
• $f''(c) < 0 \implies$ Graph is $\cap$ (concave down) → Maximum at top
• $f''(c) > 0 \implies$ Graph is $\cup$ (concave up) → Minimum at bottom
Working Rule for Local Maxima/Minima
Step 1: Find $f'(x)$ and solve $f'(x) = 0$ for critical points
Step 2: For each critical point $c$, use either:
- Method 1 (2nd Derivative Test): Find $f''(c)$ and check sign
- Method 2 (1st Derivative Test): Check sign of $f'(x)$ left/right of $c$
Step 3: Calculate $f(c)$ to find local maximum/minimum value
Absolute Maximum and Minimum Values
- Absolute Maximum: Greatest value of $f$ on entire interval $[a,b]$
- Absolute Minimum: Least value of $f$ on entire interval $[a,b]$
Working Rule (Most Important!)
Step 1: Find all critical points in $(a,b)$ where $f'(x) = 0$ or $f'(x)$ doesn't exist
Step 2: List all critical points: $c_1, c_2, c_3, ...$
Step 3: Calculate: $f(a), f(c_1), f(c_2), ..., f(b)$
Step 4: The largest value is absolute maximum
Step 5: The smallest value is absolute minimum
- Local max/min: Highest/lowest in a small neighborhood
- Absolute max/min: Highest/lowest in the entire interval
- Absolute max/min can occur at endpoints or at critical points!
Important Theorems
Theorem 2: If $f$ is differentiable and attains absolute max/min at interior point $c$, then $f'(c) = 0$.
- Strictly Increasing vs. Increasing:
• $f$ is strictly increasing on $(a, b)$ if $f'(x) > 0$ for all $x \in (a, b)$.
• $f$ is increasing on $[a, b]$ if $f'(x) \ge 0$ for all $x \in (a, b)$.
• A function can have $f'(x) = 0$ at isolated points and still be strictly increasing (e.g. $f(x) = x^3$, where $f'(0) = 0$, is strictly increasing on $\mathbb{R}$). - Critical Point Formal Definition:
A point $c$ in the domain of $f$ is a critical point if either $f'(c) = 0$ OR $f'(c)$ does not exist. - When the Second Derivative Test Fails ($f''(c) = 0$):
If $f'(c) = 0$ and $f''(c) = 0$, the test is inconclusive! You MUST revert to the First Derivative Test:
• Sign of $f'(x)$ changes from $+$ to $-$ as $x$ increases through $c$ → Local Maximum.
• Sign of $f'(x)$ changes from $-$ to $+$ as $x$ increases through $c$ → Local Minimum.
• Sign of $f'(x)$ does NOT change → Point of Inflection (neither max nor min, e.g. $f(x) = x^3$ at $x=0$). - Absolute Extrema Algorithm (Closed Interval $[a, b]$):
• Step 1: Find all critical points $c_i \in (a, b)$.
• Step 2: Compute values at critical points and boundaries: $f(a), f(c_1), \dots, f(b)$.
• Step 3: Absolute Max $= \max\{f\}$, Absolute Min $= \min\{f\}$. (Never test $f''(x)$ for absolute extrema on closed intervals!). - Rate of Change Sign Rule:
If a quantity decreases with time (e.g., ladder sliding down a wall, water leaking from a cone), its rate of change MUST be assigned a negative value: $\frac{dh}{dt} < 0$. - 📌 Syllabus Clarification: Tangents & Normals (NCERT Ex 6.3) and Approximations using differentials (NCERT Ex 6.4) have been deleted from the rationalized CBSE Class 12 syllabus.
Solved Examples (Board Pattern)
Q1. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? (2 Marks)
Let $r$ be the radius and $C$ be the circumference. Then $C = 2π r$
Differentiating w.r.t. time $t$:
$\frac{dC}{dt} = 2π \frac{dr}{dt}$
Given: $\frac{dr}{dt} = 0.7$ cm/s
Therefore: $\frac{dC}{dt} = 2π(0.7) = 1.4π$ cm/s
Answer: The circumference is increasing at $1.4π$ cm/s or approximately $4.4$ cm/s.
Q2. Find the intervals in which $f(x) = 2x^3 - 3x^2 - 36x + 7$ is (a) increasing (b) decreasing. (3 Marks)
$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)$
$f'(x) = 0 \Rightarrow x = 3$ or $x = -2$
These points divide real line into intervals: $(-∞, -2), (-2, 3), (3, ∞)$
- In $(-∞, -2)$: Take $x = -3$, $f'(-3) = 6(-6)(-5) = 180 > 0$ ? Increasing
- In $(-2, 3)$: Take $x = 0$, $f'(0) = 6(-3)(2) = -36 < 0$ ? Decreasing
- In $(3, ∞)$: Take $x = 4$, $f'(4) = 6(1)(6) = 36 > 0$ ? Increasing
(a) $f$ is increasing on $(-∞, -2) \cup (3, ∞)$
(b) $f$ is decreasing on $(-2, 3)$
Q3. Find the local maxima and local minima of $f(x) = x^3 - 3x^2 + 4x$. (3 Marks)
$f'(x) = 3x^2 - 6x + 4$
For critical points: $f'(x) = 0$
$3x^2 - 6x + 4 = 0$
Discriminant: $D = 36 - 48 = -12 < 0$
No real roots, so no critical points.
$f'(x) = 3(x^2 - 2x + \frac{4}{3}) = 3[(x-1)^2 + \frac{1}{3}] > 0$ for all $x \in \mathbb{R}$
Since $f'(x) > 0$ for all $x$, the function is strictly increasing on $\mathbb{R}$.
Therefore: No local maxima or minima exist.
Q4. Find the absolute maximum and minimum values of $f(x) = x^3$ on $[-2, 2]$. (3 Marks)
$f'(x) = 3x^2$
$f'(x) = 0 \Rightarrow x = 0$
Critical point: $x = 0$ (lies in $[-2, 2]$)
$f(-2) = (-2)^3 = -8$
$f(0) = 0^3 = 0$
$f(2) = 2^3 = 8$
Absolute Maximum: $8$ at $x = 2$
Absolute Minimum: $-8$ at $x = -2$
Word Problems & Applications
Type 1: Optimization Problems (Most Important!)
1. Read problem carefully and identify what to maximize/minimize
2. Draw diagram if needed
3. Express the quantity to be optimized as function of one variable
4. Find derivative and critical points
5. Use first or second derivative test
6. Answer in context with proper units
Classic Problem Types
Type A: Two Numbers
Problem: Find two positive numbers whose sum is $k$ and product is maximum.
Answer: Both numbers = $\frac{k}{2}$
Method: Let numbers be $x$ and $k-x$. Maximize $P = x(k-x)$
Type B: Rectangle in Circle
Problem: Rectangle of maximum area inscribed in circle of radius $r$.
Answer: Square with side $r\sqrt{2}$
Max Area: $2r^2$
Type C: Open Box
Problem: Box from rectangular sheet by cutting squares from corners.
Method: If cut $x$ cm, volume $V = x(l-2x)(b-2x)$
Find $\frac{dV}{dx} = 0$ and solve
Type D: Cylinder in Sphere
Problem: Right circular cylinder of max volume in sphere of radius $R$.
Answer: Height = $\frac{2R}{\sqrt{3}}$
Max Volume: $\frac{4π R^3}{3\sqrt{3}}$
Solved Application Problem
Q. Find two positive numbers whose sum is 16 and sum of whose cubes is minimum. (5 Marks)
Solution:
Step 1: Let the two numbers be $x$ and $16-x$ where $0 < x < 16$
Step 2: Sum of cubes: $S = x^3 + (16-x)^3$
Step 3: Differentiate: $\frac{dS}{dx} = 3x^2 - 3(16-x)^2 = 3[x^2 - (16-x)^2]$ $= 3[x^2 - 256 + 32x - x^2] = 3(32x - 256) = 96(x - 8)$
Step 4: For critical points: $\frac{dS}{dx} = 0 \Rightarrow x = 8$
Step 5: Second derivative: $\frac{d^2S}{dx^2} = 96 > 0 \implies$ Minimum at $x = 8$
Answer: The two numbers are $8$ and $8$.
Previous Year Questions (PYQs)
(A) increasing on $\mathbb{R}$
(B) decreasing on $\mathbb{R}$
(C) increasing on $(-∞, -1) \cup (1, ∞)$
(D) decreasing on $(-1, 1)$
Ans: (C) and (D) both. $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$.
Sign analysis shows increasing on $(-∞, -1) \cup (1, ∞)$ and decreasing on $(-1, 1)$.
Solution:
Volume $V = x^3$, Surface area $S = 6x^2$
Given: $\frac{dV}{dt} = 8$ cm-/s
$\frac{dV}{dt} = 3x^2\frac{dx}{dt} \Rightarrow 8 = 3(12)^2\frac{dx}{dt}$
$\frac{dx}{dt} = \frac{8}{432} = \frac{1}{54}$ cm/s
$\frac{dS}{dt} = 12x\frac{dx}{dt} = 12(12)(\frac{1}{54}) = \frac{144}{54} = \frac{8}{3}$ cm-/s
Answer: $\frac{8}{3}$ cm-/s or $2.67$ cm-/s
Solution:
$f(x) = \log(\sin x)$
$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$
In $(0, \frac{π}{2})$: $\cot x > 0$, so $f'(x) > 0$ ? increasing
In $(\frac{π}{2}, π)$: $\cot x < 0$, so $f'(x) < 0$ ? decreasing
Hence proved.
Solution Outline:
1. Let width = $2r$, height of rectangle = $h$
2. Perimeter: $2h + 2r + π r = 10$
3. Area (light): $A = 2rh + \frac{π r^2}{2}$
4. Express $h$ from perimeter: $h = \frac{10 - 2r - π r}{2}$
5. Substitute in $A$ and maximize
6. Find $\frac{dA}{dr} = 0$ and solve
Answer: Width $= \frac{20}{4+π}$ m, Height $= \frac{10}{4+π}$ m
Key Steps:
1. If cylinder has radius $r$ and height $h$, then from sphere: $r^2 + \frac{h^2}{4} = R^2$
2. Volume: $V = π r^2 h = π(R^2 - \frac{h^2}{4})h = π R^2h - \frac{π h^3}{4}$
3. $\frac{dV}{dh} = π R^2 - \frac{3π h^2}{4} = 0$
4. Solving: $h^2 = \frac{4R^2}{3} \Rightarrow h = \frac{2R}{\sqrt{3}}$
5. Verify using second derivative test: $\frac{d^2V}{dh^2} < 0$ ? Maximum
Common Mistakes & Scoring Tips
Common Mistakes
Scoring Tips
Formula Sheet (Quick Revision)
Rate of Change
1. Rate of change of $y$ w.r.t. $x$: $\frac{dy}{dx}$2. Chain Rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$ or $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
3. Marginal Cost: $MC = \frac{dC}{dx}$
4. Marginal Revenue: $MR = \frac{dR}{dx}$
Increasing/Decreasing
5. $f$ increasing on $(a,b)$ if $f'(x) > 0$ for all $x \in (a,b)$6. $f$ decreasing on $(a,b)$ if $f'(x) < 0$ for all $x \in (a,b)$
7. Critical points: Solve $f'(x) = 0$
Maxima/Minima
8. First Derivative Test:- $f'(x)$ changes from $+$ to $-$ at $c \implies$ Local maximum at $c$
- $f'(x)$ changes from $-$ to $+$ at $c \implies$ Local minimum at $c$
9. Second Derivative Test:
- If $f'(c) = 0$ and $f''(c) < 0 \implies$ Local maximum at $c$
- If $f'(c) = 0$ and $f''(c) > 0 \implies$ Local minimum at $c$
- If $f'(c) = 0$ and $f''(c) = 0 \implies$ Test fails, use first derivative test
Absolute Max/Min on [a,b]
10. Find all critical points in $(a,b)$11. Evaluate $f$ at critical points and at $a, b$
12. Largest value = Absolute maximum
13. Smallest value = Absolute minimum
Geometry Formulas
14. Circle: $A = π r^2$, $C = 2π r$15. Sphere: $V = \frac{4}{3}π r^3$, $S = 4π r^2$
16. Cone: $V = \frac{1}{3}π r^2 h$, $S = π rl$, $l = \sqrt{r^2 + h^2}$
17. Cylinder: $V = π r^2 h$, $S = 2π rh + 2π r^2$
18. Cube: $V = x^3$, $S = 6x^2$
Practice Problems (Self-Test)
Level 1: Basic (2 Marks Each)
Q1. The side of a square is increasing at the rate of 0.2 cm/s. Find the rate of increase of its perimeter.
Q2. Show that $f(x) = e^{2x}$ is increasing on $\mathbb{R}$.
Q3. Find the slope of the tangent to the curve $y = x^3 - x$ at $x = 2$.
Level 2: Intermediate (3 Marks Each)
Q4. Find the intervals in which $f(x) = x^4 - 4x^3 + 4x^2 + 15$ is increasing or decreasing.
Q5. Find all points of local maxima and minima of $f(x) = x^3 - 6x^2 + 9x + 15$.
Q6. Find the absolute maximum and minimum values of $f(x) = \sin x + \cos x$ on $[0, π]$.
Level 3: Advanced (5 Marks Each)
Q7. A wire of length 28 m is to be cut into two pieces. One piece is bent into a square and the other into a circle. What should be the lengths so that the combined area is minimum?
Hint: If square side is $x$ and circle radius is $r$, then $4x + 2π r = 28$.
Q8. Show that of all rectangles inscribed in a given fixed circle, the square has maximum area.
Q9. A tank with rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m³. If building costs Rs 70/m² for base and Rs 45/m² for sides, find the cost of the least expensive tank.
Exam Strategy & Time Management
Topic-wise Time Allocation (for 80 marks paper)
| Topic | Expected Questions | Time to Allocate | Difficulty |
|---|---|---|---|
| Rate of Change | 1 MCQ + 1 Short (2M) | 5-7 minutes | Easy |
| Increasing/Decreasing | 1 Short (3M) | 8-10 minutes | Medium |
| Local Maxima/Minima | 1 Short (3M) | 8-10 minutes | Medium |
| Absolute Max/Min or Application | 1 Long (5M) | 15-18 minutes | Hard |
Last Week Revision Strategy
Day 7-5 Before Exam
Revise all formulas and theorems
Solve 5 PYQs from each topic
Practice rate of change word problems
Master the working rules
Day 4-2 Before Exam
Focus on optimization problems
Solve full-length problems (5M each)
Revise common mistakes list
Practice mental calculation of derivatives
Last Day Before Exam
Go through formula sheet 3 times
Solve 2 MCQs, 2 Short answers, 1 Long answer
Revise the 24-hour checklist
Sleep early - fresh mind = fewer calculation errors!
1. Define variables clearly (0.5 marks)
2. Form the main equation (1 mark)
3. Differentiate correctly (1 mark)
This gives you 2.5/5 marks minimum!
Advanced Summary & Sign Testing
- If $f$ is continuous on $[a,b]$, it has both absolute max and min
- Every monotonic function attains extreme values only at endpoints
- If $f'(x) > 0$ everywhere except at isolated points where $f'(x) = 0$, then $f$ is still increasing
- Point of inflection: $f'(c) = 0$ but no extremum (e.g., $x^3$ at $x=0$)
Sign Testing Quick Reference
For $f'(x) = (x-a)(x-b)(x-c)$ where $a < b < c$:
| Interval | Sign of each factor | Sign of $f'(x)$ | Nature |
|---|---|---|---|
| $(-∞, a)$ | $(-)(-)(-)$ | $-$ | Decreasing |
| $(a, b)$ | $(+)(-)(-)$ | $+$ | Increasing |
| $(b, c)$ | $(+)(+)(-)$ | $-$ | Decreasing |
| $(c, ∞)$ | $(+)(+)(+)$ | $+$ | Increasing |
Chapter 6 Board Mastery Test
Practice test • 10 marks • 25 min